Linear First Order ODE/y' = x + y

Theorem

$(1): \quad \dfrac {\d y} {\d x} = x + y$

has the general solution:

$y = C e^x - x - 1$

Initial Value: $\map y 0 = 1$

$(1): \quad \dfrac {\d y} {\d x} = x + y$

with initial condition:

$\map y 0 = 1$

has the particular solution:

$y = 2 e^x - x - 1$

Proof

Rearranging $(1)$:

$(2): \quad \dfrac {\d y} {\d x} - y = x$

$(2)$ is a linear first order ODE in the form:

$\dfrac {\d y} {\d x} + \map P x y = \map Q x$

where:

$\map P x = -1$
$\map Q x = x$

Thus:

 $\displaystyle \int \map P x \rd x$ $=$ $\displaystyle -\int \rd x$ $\displaystyle$ $=$ $\displaystyle - x$ $\displaystyle \leadsto \ \$ $\displaystyle e^{\int P \rd x}$ $=$ $\displaystyle e^{-x}$ $\displaystyle$ $=$ $\displaystyle \frac 1 {e^x}$

Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:

 $\displaystyle \map {\dfrac {\d} {\d x} } {\dfrac y {e^x} }$ $=$ $\displaystyle \frac x {e^x}$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac y {e^x}$ $=$ $\displaystyle \int \frac x {e^x} \rd x$ $\displaystyle$ $=$ $\displaystyle -\frac 1 {e^x} \paren {x + 1} + C$ Primitive of $x e^{a x}$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle C e^x - x - 1$

$\blacksquare$